### PROBLEM LINK:

**Author:** admin

**Tester:** Kevin Atienza

**Editorialist:** Ajay K. Verma

**Russian Translator:** Sergey Kulik

**Mandarian Translator:** Gedi Zheng

### DIFFICULTY:

Medium

### PREREQUISITES:

Combinatorics, Modular Arithmetic

### PROBLEM:

Given two integers N and K, find how many triplets of strings over K alphabets exist, such that length of all strings is bounded by N, and no string in a triplet is a prefix of another string of the triplet.

### EXPLANATION:

We compute the number of triplets of strings of bounded length, and then subtract the ones in which one string is a prefix of another. For the sake of brevity, we define a new notation \prec to represent the “prefix” relationship, i.e., "X is a prefix of Y" can be represented by X \prec Y.

##Total Number of Triplets:

There are exactly K^i strings of length i. Hence, the number of strings with length not exceeding N will be:

A = K + K^2 + K^3 + \cdots + K^N

A = K (K^N - 1 / (K - 1))

If the modular inverse of (K - 1) with respect to the given prime p = 10^9 + 7 is x, then the value of A \bmod p would be (K (K^N - 1) x) \bmod p.

Both the modular inverse, and power of a number can be compute in O (\log N) time. Hence, we can compute the value of A in O (\log N) time.

There are A strings of length not exceeding N. Therefore, the number of triplets would be A^3. Next, we count how many of these triplets are bad triplets, i.e., the ones in which one string is a prefix of another.

##Triplets of the Form X \prec Y \prec Z:

In this section we count the number of triplets which have strings forming a chain under the relationship \prec, i.e., if the triplet is (P, Q, R), then one of the following is true:

P \prec Q \prec R

P \prec R \prec Q

Q \prec P \prec R

Q \prec R \prec P

R \prec P \prec Q

R \prec Q \prec P

Note that for some triplets more than one of these conditions might be true, e.g., (X, X, X) will satisfy all 6 conditions. Hence, we consider four categories, so that each triplet is counted exactly once.

**1) X = Y = Z:**

There will be A triplets of this form, as we can choose the string X in A ways, and the other two will be equal to it. These triplets will satisfy all 6 conditions, i.e., each triplet will be repeated 6 times.

**2) X \prec Y = Z and X
eq Y:**

For a given string Y of length r, there are exactly (r - 1) possible values of string X, the ones which are proper substring of Y. Hence, the number of such triplets would be:

K^2 + 2 K^3 + 3 K^4 + \cdots + (N - 1) K^N.

This is the sum of an arithmetic-geometric series, and can be represented explicitly as shown here. The computation of this sum also requires the computation of power and modular inverse, and hence can be done in O (\log N) time.

These triplets will satisfy two of the above conditions (X \prec Y \prec Z, and X \prec Z \prec Y).

**3) X = Y \prec Z, and Y
eq Z:**

Number of these triplets will be the same as the one computed in (2). This is because once we pick a string Z of length r, then Y has exactly (r - 1) choices.

These triplets also satisfy two of the above conditions (X \prec Y \prec Z, and Y \prec X \prec Z).

**4) X \prec Y \prec Z, X
eq Y, and Y
eq Z:**

If we pick a string Z of length r, then we only need to choose the lengths of X and Y. The lengths of X and Y must be distinct and lie between 1 and (r - 1). Hence, the number of (X, Y) pairs would be the number of ways of choosing two distinct numbers between 1 and (r - 1), which is given by (r - 1) \choose 2. Therefore the number of such triplets would be:

K^3 + 3 K^4 + 6 K^5 + \cdots + {{N - 1} \choose 2} K^N

Again this is the sum of arithmetic-geometric sequence of second order, and can be computed using the similar methods. The computation will take O (\log N) time.

These triplets will satisfy exactly one of the above conditions.

Hence, the number of chain triplets would be 6 ((number of category 1 triplets)/6 + (number of category 2 triplets)/2 + (number of category 3 triplets)/2 + (number of category 4 triplets)/1).

##Triplets of the Form "X \prec Y, Z does not share prefix relationship with Y":

In this section we count the number of triplets in which exactly two strings satisfy the “prefix” relationship, i.e., if the triplet is (P, Q, R), then one of the following is true:

P \prec Q, R has no prefix relation with Q,

Q \prec P, R has no prefix relation with P,

P \prec R, Q has no prefix relation with R,

R \prec P, Q has no prefix relation with P,

Q \prec R, P has no prefix relation with R,

R \prec Q, P has no prefix relation with Q,

Once again it is possible that one triplet satisfy more than one condition, however, we should count it only once.

**1) X = Y, Z has no prefix relation with Y:**

If we have a string Y of length r, then there are exactly r prefix strings of Y. If a string W has Y as a proper prefix, then W can be written as W = Y + U, where U is a non-empty string of length at most (N - r). Hence, the number of such string W will be K + K^2 + \cdots + K^{N - r}. Since Z is neither a prefix of Y, nor has Y as its prefix, the number of possible values of Z would be:

(K + K^2 + \cdots + K^N) - (K + K^2 + \cdots + K^{N - r}) - r

= (K^{N - r + 1} + K^{N - r + 2} + \cdots + K^N - r).

Also number of ways of choosing a r length string Y is K^r. Hence, the number of such triplets (X, Y, Z), with Y being a r-length string will be:

K^r (K^{N - r + 1} + K^{N - r + 2} + \cdots + K^N - r)

= K^{N + 1} * (1 + K + K^2 + \cdots + K^{r - 1}) - r * K^r

If we take the sum of this expression over all possible values of r, we get the number of triplets (X, Y, Z) satisfying the above criteria, and it will be:

K^{N + 1} (N + (N - 1) K + (N - 2) K^2 + \cdots + K^{N - 1}) - (K + 2 K^2 + 3 K^3 + \cdots + N K^N)

This is also a sum of arithmetic-geometric sequence, and can be computed in logarithmic time. These triplets satisfy two of the above conditions (X \prec Y, with Z having no relationship, and Y \prec X, with Z having no relationship).

**2) X \prec Y, X
eq Y, and Z has no prefix relation with Y:**

Once again if we pick a string Y of length r, then X will have (r - 1) choices, and Z will have (K^{N - r + 1} + K^{N - r + 2} + \cdots + K^N - r) choices. Hence, the number of such triplets with Y being of length r would be:

r K^r (K^{N - r + 1} + K^{N - r + 2} + \cdots + K^N - r)

= r K^{N + 1} (1 + K + K^2 + \cdots + K^{r - 1}) - r^2 K^r

If we sum it over all possible values of r, we would get the number of triplets (X, Y, Z) satisfying the above condition, which would be (after some simplifications):

{(N + 1) \choose 2} K^{N + 1} (1 + K + K^2 + \cdots + K^N) - K^{N + 1} ({2 \choose 2} K + {3 \choose 2} K^2 + \cdots + {N \choose 2} K^{N - 1})

- (1^2 K + 2^2 K^2 + 3^2 K^3 + \cdots + N^2 K^N).

The number of triplets where exactly two strings satisfy the “prefix” relationship will be 6 ( (number of triplets of first category)/2 + (number of triplets of second category)/1).

Now that we have computed the number of bad triplets, we can subtract them from the total number of triplets to get the number of good ones.

##Time Complexity:

O (\log N)